Thread: computing the slope of the secant line between points

Hello everyone. I am really getting lost trying to figure out how to correctly figure this one out. Here is the problem:

Compute the slope of the secant line between the points at (a) x = 1 and x = 2, (b) x = 2 and x = 3, (c) x = 1.5 and x = 2, and (d) x = 2 and x = 2.5, (e) x = 1.9 and x = 2, (f) x = 2 and x = 2.1, and (g) use parts (a) - (f) and other calculations as needed to estimate the slope of the tangent line at x = 2.....

f(x) = e^x

To give some context here, I have been learning (In Calculus: Early Transcendental Functions) about Differentiation, and specifically tangent lines and velocity.

Any help with this problem would be greatly appreciated.

Thanks in advance!

The point of the problem is that as the second point approaches x=2 (from both directions), the secant line approaches the tangent line.

What is a secant line? You just take two points on the graph and draw a line through them. So in (a), for example, you calculate f(1) and f(2) which are your y-values, and then just calculate the slope as:



- Hollywood

Thanks Hollywood, I understand it a lot better now, however I do not quite understand how I am supposed to calculate f(1) and f(2), for example in (a), because the problem has f(x) = e^x , So what am I supposed to do with the "e" or have I missed something somewhere?

From what I gathered, to find f(x) = e^x , replace the x with 1,which gives us f(x) = e¹ , so what now?

Thanks in advance!!

The symbol e just represents a number . You probably have the function on your calculator.

So is just a number, and so is , , etc.

So for part (a), for example, the slope is:



- Hollywood

Got ya, thanks Hollywood!!!